What is the solution of $\sqrt{x-4}+5=2$ ?
We are given the algebraic equation \(\sqrt{x-4} + 5 = 2\). Our objective is to determine the value of \(x\) that satisfies this equation. The primary constraint here is the nature of the square root function, which is defined to produce only non-negative outputs for any valid input in its domain. We need to isolate the radical expression and see if a valid solution exists.
To isolate the radical expression, we subtract 5 from both sides of the equation:
\[\sqrt{x-4} + 5 - 5 = 2 - 5\]
\[\sqrt{x-4} = -3\]
By definition, the principal square root function \(\sqrt{u}\) returns a non-negative real value for any \(u \ge 0\). Specifically, \(\sqrt{u} \geq 0\). In our isolated equation, we have \(\sqrt{x-4} = -3\). Since the square root of any real number must be greater than or equal to zero, it is impossible for the square root to equal a negative number like -3.
Because the equation \(\sqrt{x-4} = -3\) requires the result of a square root to be negative, which contradicts the fundamental properties of the square root function, there is no real value of \(x\) that satisfies the original equation. Therefore, the correct conclusion is that there is no solution.
This problem demonstrates why it is essential to consider the domain and range of mathematical functions when solving equations. For example, in engineering or physics, if you are modeling the distance an object travels as a function of time (where distance must be positive), you might encounter equations that seem to have solutions algebraically but are physically impossible. Recognizing that a square root cannot be negative prevents one from incorrectly assigning "solutions" that do not actually exist in the physical or real-number domain.